Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int g[100][100];
int n, m;
vector<int> col;
vector<int> row;
void printG() {
  for (int j = 0; j < n; j += 1) {
    for (int i = 0; i < m; i += 1) cout << g[i][j] << " ";
    cout << endl;
  }
}
int scanVer() {
  for (int i = 0; i < m; i += 1) {
    bool flag = false;
    for (int j = 0; j < n; j += 1) {
      if (g[i][j] == 0) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      return i;
    }
  }
  return -1;
}
int scanHor() {
  for (int i = 0; i < n; i += 1) {
    bool flag = false;
    for (int j = 0; j < m; j += 1) {
      if (g[j][i] == 0) {
        flag = true;
        break;
      }
    }
    if (!flag) {
      return i;
    }
  }
  return -1;
}
void printCR() {
  int col_size = col.size();
  int row_size = row.size();
  cout << col_size + row_size << endl;
  for (int i = 0; i < col_size; i++) {
    cout << "col " << col.at(i) << endl;
  }
  for (int i = 0; i < row_size; i++) {
    cout << "row " << row.at(i) << endl;
  }
}
void ans() {
  if (n > m) {
    int res = scanVer();
    while (res != -1) {
      col.push_back(res + 1);
      for (int i = 0; i < n; i += 1) {
        g[res][i]--;
      }
      res = scanVer();
    }
    res = scanHor();
    while (res != -1) {
      row.push_back(res + 1);
      for (int i = 0; i < m; i += 1) {
        g[i][res]--;
      }
      res = scanHor();
    }
    int flag = true;
    for (int i = 0; i < n; i += 1) {
      for (int j = 0; j < m; j += 1) {
        if (g[j][i] != 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag)
      printCR();
    else
      cout << "-1" << endl;
  } else {
    int res = scanHor();
    while (res != -1) {
      row.push_back(res + 1);
      for (int i = 0; i < m; i += 1) {
        g[i][res]--;
      }
      res = scanHor();
    }
    res = scanVer();
    while (res != -1) {
      col.push_back(res + 1);
      for (int i = 0; i < n; i += 1) {
        g[res][i]--;
      }
      res = scanVer();
    }
    res = scanHor();
    while (res != -1) {
      row.push_back(res + 1);
      for (int i = 0; i < m; i += 1) {
        g[i][res]--;
      }
      res = scanHor();
    }
    int flag = true;
    for (int i = 0; i < n; i += 1) {
      for (int j = 0; j < m; j += 1) {
        if (g[j][i] != 0) {
          flag = false;
          break;
        }
      }
    }
    if (flag)
      printCR();
    else
      cout << "-1" << endl;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i += 1)
    for (int j = 0; j < m; j += 1) cin >> g[j][i];
  ans();
  return 0;
}